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If $3^n>n$ then $3^{n+1}>n+1$

I could not solve this problem

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    What have you tried so far? Is at least the result intuitively clear to you? Do you know proving statements by induction? – 2017-01-26
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    @Imago I know that first we must show P(1) is true, 2nd step we must assume P(k). Then final step we must show it is true for P(k+1). Then only it will be proved. But i could not understand this question can you explain me. – 2017-01-27

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If $n=0$, then the truth of this statement is easily verifiable, since

$$3^0 = 1 > 0.$$

If $n>0$, i.e. $n\ge1$, then:

$$3^{n+1} = (by \: basic \: arithmetic)$$

$$3\cdot3^n > (by \: assumption \: 3^n > n)$$

$$3n = (by \: basic \: arithmetic)$$

$$2n + n \ge (by \: n \ge 1)$$

$$2n + 1 > (by \: n \ge 1)$$

$$n + 1$$

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    Can you please explain the answer by writing each line in such a way which would be easy to understand. – 2017-01-27
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    @bappy Sure .. give me a sec – 2017-01-27
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Presumably, $n\in\Bbb N$ since we're talking about an induction proof (you could also include $n=0$ but I won't here). Thus, $$n\geq 1\implies n+n\geq n+1\implies n+n+n>n+1\implies 3n>n+1$$ You also have $3^n>n$ by assumption. I'll leave the rest to you.